?$p_{解}：$??$(1)$?∵拋物線?$y=-x^2+mx+3$?過點?$B(3，$??$0)$?
∴?$-9+3\ \mathrm {m}+3=0$?
∴?$m=2$?
∴拋物線的表達式為?$y=-x^2+2x+3$?
?$(2) $?由?$\begin{cases}{y=-x^2+2x+3}\\{y=- \dfrac {3}{2} x+3}\end{cases}，$?解得?$\begin{cases}{x=0}\\{y=3}\end{cases}，$?解得?$\begin{cases}{x=\dfrac 72}\\{y=- \dfrac {9}{4}}\end{cases}$?
∴?$C(0，$??$3)，$??$D(\frac {7}{2}，$??$- \frac {9}{4} )$?
∵?$S_{△ABP}=4S_{△ABD}$?
∴?$\frac {1}{2}\ \mathrm {AB} ×|y_P| =4× \frac {1}{2}\ \mathrm {AB}× \frac {9}{4}$?
∴?$|y_P| =9，$?即?$y_P=±9$?
當?$y=9$?時，?$-x^2+2x+3=9$?
∴?$x^2-2x+6=0$?
∵根的判別式為?$(-2)^2-4×1×6< 0$?
∴此方程無實數(shù)解
當?$y=-9$?時，?$-x^2+2x+3=-9$?
解得?$×1=1+ \sqrt{13}，$??$×2=1- \sqrt{13}$?
∴?$P(1+\sqrt{13}，$??$-9)$?或?$P(1- \sqrt{13}，$??$-9)$?
?$ (3) $?由?$y=-x^2+2x+3$?
令?$y=0，$?得?$-x^2+2x+3=0$?
解得?$×1=-1，$??$×2=3$?
∴?$A(-1，$??$0)，$??$B(3，$??$0)$?
拋物線的對稱軸為直線?$x=-\frac 2{2×(-1)} =1$?
當?$P、$??$D$?兩點關于拋物線的對稱軸對稱時，滿足條件，此時?$P (-\frac {3}{2} ，$??$-\frac {9}{4})$?
過點?$B$?作?$BP'//AD$?交拋物線于點?$P'，$?此時?$P '$?滿足條件
由?$A(-1，$??$0)，$??$D(\frac {7}{2} ，$??$-\frac {9}{4} )$?
可得直線?$AD$?的表達式為?$y=- \frac {1}{2} x- \frac {1}{2}$?
∵?$B(3，$??$0)，$??$BP'//AD$?
∴直線?$BP'$?的表達式為?$y=- \frac 12x+\frac 32$?
由?$\begin{cases}{y=-\dfrac 12x+\dfrac 32}\\{y=-x^2+2x+3}\end{cases}，$?解得?$\begin{cases}{x=3}\\{y=0}\end{cases}，$?或?$\begin{cases}{x=-\dfrac {1}{2}}\\{y=\dfrac {7}{4}}\end{cases}$?
∴?$P'(-\frac {1}{2} ，$??$\frac {7}{4})$?
綜上所述，滿足條件的點?$P $?的坐標為?$(-\frac {3}{2}，$??$- \frac {9}{4} $?或?$(-\frac {1}{2}，$??$ \frac {7}{4} )$?
